Infrared Spectroscopy and Mass Spectroscopy Chemistry Exam Practice

Infrared Spectroscopy andMass Spectroscopy
Dr. Niki Shoup
CHEM 3411—Summer 2021
What is Spectroscopy
• We’ve have spent this whole semester discussing chemical theory and
mechanisms
• All of the reactions I have taught you and the practice problems you
have done are from real examples of chemistry that happens in flasks
• So the next logical step in our discussion is learning about how we
identify what we have made
• This is where spectroscopy comes in!
What is Spectroscopy
• Spectroscopy takes advantage of the interactions between light and
matter
• Electromagnetic radiation
• Light is both a wave and a particle
• When it is a particle, we call this a photon!
• The wave properties of light include wavelength and frequency
• Wavelength is inversely proportional to energy
• Frequency is directly proportional to energy
What is Spectroscopy
What is Spectroscopy
What is Spectroscopy
• When we scale everything down from the macro scale to the micro
scale, we have to consider the quantum behavior of matter
• This means that matter experiences quantum behavior
• Quantum is a fancy way of saying discrete numbers
• Matter also has wave like properties which is why light can interact
with it!
• We take advantage of this for infrared (IR) spectroscopy
IR Spectroscopy
• IR spectroscopy measures the bond vibrational frequencies of
molecules
• This works by using IR light to excite the bond and causes the bonds
to vibrate more intensely
• The vibrational modes observed are:
IR Spectroscopy
• Let’s look at some bond energy gaps and relate them to vibrational
energy
IR Spectroscopy
How does and IR spectrometer work?
• The sample is placed on a plate (or, in the old times) dissolved in a
solvent or embedded in a KBr pellet.
• The sample is then irradiated with IR light
• The frequencies that are absorbed will cause bonds to vibrate faster!
• This can then tell us which functional groups are present!
IR Spectroscopy
• We look at an IR spectrum that plots the %transmittance (how much
light got though) vs the frequency in wavenumbers
IR Spectroscopy
• Wavenumbers are directly
proportional to energy
• The values we typically see
are between 400-4000 cm-1
• The three important
characteristics of an IR signal
are:
• Wavenumber
• Intensity
• Shape
Wavenumbers!
• The frequency of a stretching vibration depends on the bond strength and
the mass difference of the atoms
• The stronger the bond the higher the stretching frequency
• The larger the mass difference the higher the stretching frequency
Wavenumbers!
• So we can come up with some trends!
Wavenumbers!
• The region above 1500 cm-1 is called the diagnostic region
• The signals in this region give clear information on the functional
groups present
• Below 1500 cm-1 is the fingerprint region, which can be hard to
analyze
Wavenumbers
• We can compare the IR stretching of carbon-hydrogen bonds in order
to get information on the type of hybridization the carbon atom has
• Them more s-character of the carbon, the stronger the C-H bond, so
the higher frequency is observed.
Wavenumbers
• We can actually see this cut off on the IR
• (I think its neat!)
Wavenumbers
• Carbonyl compounds have a
specific wavenumber signals
between 1800 and 1600 cm1
• These signals can be affected
by other functionality near
the carbonyl
Intensity
• The strength of the signal is called the intensity
Intensity
• When bonds undergo stretching vibrations, the dipole moment will also oscillate
• This will create an electric field around the bond
• The more polar the bond the greater the interaction between the dipole and the
IR radiation
Intensity
• Note the difference between
and alkene and a carbonyl
• And if we have symmetrical
bonds, we don’t see the signal
Intensity
• Stronger signals are also observed when there are multiple bonds of
the same type vibrating
• Although C-H bonds are not very polar, they often give very strong
signals, because there are often many of them in an organic
compound
• Because sample concentration can affect signal strength, the
Intoxilyzer 5000 can be used to determine blood alcohol levels be
analyzing the strength of C-H bond stretching in blood samples
Shape
• Some signals are broad, others are narrow
Shape
• Signals for alcohols and carboxylic acid are quite broad because of
hydrogen bonding in the system
• Because H-bonds are transient IMAFs we will see compounds we
different H-bond strength
• If the sample is dilute, or if we have an independent alcohol (that
cannot hydrogen bond) our OH stretch will be narrow.
Shape
• Because of alcohols ability to form H-bonds we will see several signals
due to H-bonding
Shape
• Extensive hydrogen bonding (like what we see for a carboxylic acid)
will give rise to very large stretching that overtakes part of your IR
spectrum
Shape
• Primary and secondary
amines have N-H stretching
signals that are IR active
• These stretches are a
broadened (not that sharp)
depending on how much Hbonding is present.
• Primary amines have two
signals
• Secondary amines have one
signal
Shape
• For primary amines we see two signals because we
see both symmetric and asymmetric stretching.
• Because at any one time half the molecules are
stretching symmetrically while the other molecules
are stretching asymmetrically, we will see both.
Analyzing an IR Spectrum
• We us IR to help us identify the functional groups in a compound
• To do this we break the spectrum down into the regions above 1500
cm-1
a) 1600-1850 cm-1 – check for double bonds
b) 2100-2300 cm-1 – check for triple bonds
c) 2700-4000 cm-1 – check for X-H bonds
d) Analyze wavenumber, intensity, and shape for each signal
Analyzing an IR Spectrum
• For the 2700-4000 cm-1 region
draw a line at 3000 cm-1 so that
you can determine which
functional group is generating
the signal
Analyzing an IR Spectrum
• Because IR allows us to determine which functional groups present,
we can use it as a clue if a reaction is successful
• The oxidation of cyclohexanol to cyclohexanone we can use IR to
determine if the reaction is working
• The loss of the alcohol at 3100 cm-1 and the appearance of a carbonyl at
1700 cm-1 would be an excellent clue!
IR Spectroscopy—Limitations
• IR can only tell us about which functional groups are present
• So compounds like 2-butanol and 2-propanol have very similar IRs
Examples
Examples
Mass Spectrometry
• Mass spectrometry is used to determine the molar mass and formula
for a compound
• This works by:
• Vaporized the compound, which is then ionized, and undergoes
fragmentation
• The mass of the separate ions are detected and graphed
Mass Spectrometry
• Mass spec is one of our most sensitive and widely used techniques!
• Many organic compounds can be identified
• Pharmaceutical: drug discovery and drug metabolism, reaction monitoring
• Biotech: amino acid sequencing, analysis of macromolecules
• Clinical: neonatal screening, hemoglobin analysis
• Environmental: drug testing, water quality, food contamination testing
• Geological: evaluating oil composition
• Forensic: Explosive detection
• Many More
Mass Spectrometry
• The most common method of ionizing molecules is by electron
impact
• This works by bombarding the sample with high energy electrons
(1600 kcal or 70 eV)
• This will cause an electron to be ejected from the molecule
Mass Spectrometry
• The mass of the radical cation is the same as that of the parent
compound
• Electrons are really tiny!
• The radical cation is labeled as (M+.)
• The radical cation will then fragment into a radical and a cation
Mass Spectrometry
• The mass spectrum is a graph of the
relative abundance of each cation that is
detected.
• The x-axis is the mass to charge ratio (m/z)
which is the mass of the cation
• We do not see the mass of the fragmented
radicals
• The base peak is the tallest peaks
• We make this 100% relative abundance
• This may or may not be the parent ion
Mass Spectrometry
• Let’s look at how methane
fragments
• Each subsequent fragment is a
hydrogen radical leaving.
Analyzing the Base Peak!
• Let’s start with benzene
• The base peak is benzene
itself
• and it doesn’t fragment
easily
Analyzing the Base Peak
• For pentane we see a
different story
• Pentane (and a lot of
alkanes) will easily fragment
making the base peak the
most stable radical.
Analyzing the Base Peak!
• The first step in analyzing a mass spec is to identify the M+• peak
• The m/z of the parent ion = molar mass of the compound
• An odd massed M+• peak generally means there is an odd number of N atoms
in the molecule
• An even massed M+• peak generally indicates an absence of nitrogen, or an
even number of N atoms are present
Analyzing the
.
+
(M+1)
peak
• For several compounds we will see
an (M+1) peak
• This is due to the relative
abundance of carbon 13 (1%) we
will see it in the mass spec
Analyzing Isotopes Using Mass Spec
• Chlorine has two abundant
isotopes: 35Cl=76% and
37Cl=24%
• We can see them in the mass
spec
• If you can see carbon 13 you can
see pretty much all of the other
abundant isotopes
• That means compounds with
chlorine will see an M and a
M+2 peak with a 3:1 ratio
Analyzing Isotopes Using Mass Spec
• Bromine has two common
isotopes: 79Br=51% and
81Br=49%
• That means we will see an M
and a M+2 peak of relatively
equal intensity
Analyzing Fragments!
• As I’ve talked about, we generate radicals during EI
• The radicals will break apart and give us several fragments
• Let’s look at pentane
Analyzing Fragments
• The clusters next to the mass peaks are the hydrogens fragmenting
off
Analyzing Fragments
• Compound will continue to follow chemistry even while we are doing
mass spec on them
• So they will fragment to for the most stable carbocation
• We can see other fragmentations (and we do record them!) the
largest signals are from stable fragments.
Analyzing Fragments
• Several fragments can rearrange to form more stable radical cations
A stabilized oxonium
ion is formed
A stable, neutral
molecule is formed
Analyzing Fragments
• Amines undergo alpha cleavage
• Carbonyls undergo McLafferty rearagement
Analyzing Fragments
• Common mass fragments
High Resolution Mass Spec
• High Resolution Mass Spectrometry allows m/z to be measured with
up to 4 decimal places
• Masses are generally not whole number integers
• 1 proton = 1.0073 amu and 1 neutron = 1.0086 amu
• One 12C atom = exactly 12.0000 amu, because the amu scale is based
on the mass of 12C
• All atoms other than 12C will have a mass in amu that can be
measured to 4 decimal places by a high-resolution mass spec
instrument
High Resolution Mass Spec
• High resolution mass spec allows us to
be able to distinguish between
compounds with the same molecular
weight by looking at the relative
abundance difference of isotopes and
slight differences in AMU
84.0573 amu
84.0936 amu
GC-MS
• Mass spectroscopy is an identification technique and is best suited for pure substances.
• By using a MS as the final detector of a chromatography system we can increase the
power of both the separation technique and the identification of pure substances.
• The most common is gas chromatography
GC-MS
• GC—separates the components in a mixture
• Each of these signals is then put though a MS for mass analysis and
identification.
MS of Large/Sensitive Compounds
• For large or sensitive compounds EI will cause fragmentation and is
not a good technique for compounds that cannot be easily vaporized
(put into gas form)
• Electrospray ionization (ESI) is a soft technique in which a highvoltage needle sprays liquid solution of the analyte into vacuum
Degrees of Unsaturation
• We know that saturated hydrocarbons have the following formula
CnH2n+2
• We can use MS to help us determine if a compound deviates from
that formula by using the hydrogen deficiency index (HDI)
• For every degree of unsaturation ( a double bond or a ring) you will
lose two hydrogens
Degrees of Unsaturation
• Consider the isomers of C4H6
• All of the options have 2 degrees of unsaturation!
• So each compounds HDI = 2
Putting the Techniques Together
• IR is a technique used to determine the functional groups
• MS is used to determine the molar mass and can help us get the
molecular formula
• By using these in tandem we can identify a compound!
Example!
OH
Question 1
When you perform hydroboration-oxidation the alcohol ends up on the
O on both positions
O Markovnikov position
O anti-Markovnikov position
O no product is formed
D
Question 2
1 pt
Halohydrin formation gives you the
O anti-addition product
19
**
tv
MacBook Air
80
888
a
FS
FO
ליז
DI
FB
D
o
A
4
%
5
&
7
C
.
6
00 *
9
R
T
Y
U
1
o
F
G
I
J
к
V
B
N
c
M
C 89
IN
ucdenver.instructure.com/courses/460738/quizzes/secured#lockdown
uiz Tools
Question 2
1 pts
Halohydrin formation gives you the
anti-addition product
syn-addition product
00%
exclusively R products
se
O exclusively S products
D
Question 3
1 pts
Oxidative cleavage gives you
O a carbon-carbon double bond
O two halogens
O two carbonyls
two alcohols
Question 4
1 pts
19
tv
MacBook Air
So
ga
F&
CIN
F
F9
10
3
4
5
(
6
&
7
8
9
)
O
R
T
Y
U
I
0
P
F
G
H
J
к
L
C
V
B
N
< M Quiz Tools Question 4 1 pts Consider the following reaction sequence; what steps did you take? OH O radical bromination, Hoffman elimination, hydroboration-oxidaiton 100% radical bromination, Zaitsev elimination, oxymercuration-demercuration pse OHBr addition, Zaitsev elimination, oxymercuration-demercuration O HBr addition, Hoffman elimination, hydroboration-oxidation D Question 5 1 pts The step in radical chemistry that gives your balanced chemical equation is... O termination O all three O propagation initiation 19 étv & MacBook Air 3 888 Fa . DI DO Es $ 2 VO A 3 5 6 & 7 8 9 0 E R T Y U 1 0 Р F G H J к C V B N. M 100% - E 9:00 IN ucdenver.Instructure.com/courses/460738/quizzes/secureditlockdown D Question 6 >>
1 pts
Using osmium to make diols will happen in a…
syn-addition
O anti-addition
OS-configuration
OR-configuration
D
Question 7
1 pts
Adding HBr and peroxide to a compound containing a double bond will give hydrohalogenation
will lead to…
anti-Markovnikov addition
no product is formed
Markovnikov addition
O syn-addion
D
Question 8
1 pts
19
F
atv A
MacBook Air
90
888
ES
+7
DII
F8
مي
F10
FM
3
$
4
RLO
6
7
(
9
7
8
0
1
=
E
R
T
Y
U
1
0
Р
F
G
H
J
к
L
с
V
B
N
M
V –
100%
ucdenver.instructure.com/courses/460738/quizzes/secured lockdown
– E’T 9:00 a
D
Question 8
1 pts
Which of the following is a list of halogen radicals from least reactive to most reactive?
O Cl